Integrate a finite current segment before using an infinite limit

Carry the source element direction

A straight segment runs from x = −a to +a with current I toward +x, and the observer is (0,y,0), y positive. The source displacement is (−x,y,0). Biot–Savart gives dB = mu0 I(dl cross r)/(4pi r³); dl cross r points +z and has magnitude y dx. Thus Bz = mu0 I/(4pi) integral y dx/(x² + y²)^(3/2).

Evaluate the definite integral

An antiderivative is x/[y sqrt(x² + y²)]. Evaluating symmetric endpoints gives Bz = mu0 I a/[2pi y sqrt(a² + y²)]. The direction follows the right-hand cross product, not merely the positive magnitude formula. A small source segment has no field component along itself or toward the observation point; the contribution is tangential around its current direction.

Take a controlled limit

For a much larger than y, a/sqrt(a² + y²) approaches one, yielding the long-wire result mu0 I/(2pi y). Current is held fixed in that limit. At comparable a and y, replacing the finite source by the infinite result overestimates its contribution. Midpoint sums provide a numerical integral check, but their agreement is not independent experimental validation.

Complete the physical source account

A real steady current cannot simply end at an isolated segment. This calculation is the contribution of one portion of a larger closed circuit; omitted return-path contributions must be negligible or separately calculated. Moving the return conductor close to the observer can invalidate the simple segment-only inference.

Worked original case

For a = y = 1 m and I = 2 A, Bz ≈ 0.282843 microtesla. The long-wire comparison is 0.4 microtesla, so the ratio is 1/sqrt(2). The 64-strip midpoint approximation is closer than the 8-strip result for this smooth supported packet.

Investigation

Compare source half-length and observation distance separately, retaining the exact ratio and two numerical resolutions. Sketch current, source displacement and cross-product direction for one negative-x and one positive-x element. Propose an observation that could reveal an omitted nearby return path.

Changed-case practice

Try a new practice case

If a/y = sqrt(3), what is the finite-to-infinite field ratio?

Numerical result in the requested units Check this practice result

This public, unsaved check is not a secure assessment. Explain the source integral, symmetry or charge boundary in the retained course evidence workspace.

Sources and model boundary

Original teaching checked 31 October 2026 against the current official framework, Unit 12. The displayed magnetic calculations use the stipulated introductory approximation mu0 = 4pi×10⁻⁷ SI, not a claim that this is exact in modern SI. All apparatus and records are virtual, supplied or generated. No real magnet handling, electrical current setup, particle apparatus or material operation is requested. Public feedback and retained reports are practice evidence, not independent mastery or performed laboratory time.

Supplemental primary teaching reference: The Biot–Savart law. The supplied cases, derivations and diagrams here are original; the source’s older exact-mu0 convention and classical spin language are not adopted literally.